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lara [203]
3 years ago
15

If the graph of y = |x| is translated so that the point (1, 1) is moved to (4, 1), what is the equation of the new graph?

Mathematics
1 answer:
Kipish [7]3 years ago
4 0
First you graph all of these answers in a calculator:

y= Ix-3I is correct, it moved to (4,1).

y=IxI-3 is incorrect because it moved to (1,-2)

y=Ix +3I is incorrect because it moved to (-2,1)

y=IxI+3 is incorrect because it moved to (1,4)

So your answer is A: y= Ix-3I
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Answer: 13

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3 years ago
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A recent survey found that 7575​% of all adults over 50 wear glasses for driving. In a random sample of 4040 adults over​ 50, wh
malfutka [58]

Answer:

Mean = 30

standard Deviation = 2.74

Step-by-step explanation:

Formula

mean = np

Standard Deviation = \sqrt{npq}

from the question = 40

P = 75% or 0.75

q = 1 - p = 1 - 75% or 1 - 0.75

To find mean

mean = 40 x 0.75

           = 30

For Standard Deviation (SD)

Standard Deviation = \sqrt{npq}

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5 0
3 years ago
Please answers all these questions for me need help too pass plz help☺
vodka [1.7K]

Please take 1 minute, and really read the example solution
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8 0
3 years ago
Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i
rosijanka [135]

It looks like the ODE is

y'+5y=\begin{cases}0&\text{for }0\le t

with the initial condition of y(0)=4.

Rewrite the right side in terms of the unit step function,

u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t

In this case, we have

\begin{cases}0&\text{for }0\le t

The Laplace transform of the step function is easy to compute:

\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s

So, taking the Laplace transform of both sides of the ODE, we get

sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s

Solve for Y(s):

(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}

We can split the first term into partial fractions:

\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs

If s=0, then 1=5a\implies a=\frac15.

If s=-5, then 1=-5b\implies b=-\frac15.

\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}

\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}

Take the inverse transform of both sides, recalling that

Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)

where F(s) is the Laplace transform of the function f(t). We have

F(s)=\dfrac1s\implies f(t)=1

F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}

We then end up with

y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}

3 0
4 years ago
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Ilya [14]

Answer:

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